A plano-convex lens behaves as a concave mirror, if its one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur. Let the focal length of silvered lens be F.
$$
\begin{aligned}
\frac{1}{F} &=\frac{1}{f}+\frac{1}{f}+\frac{1}{f_{m}} \\
&=\frac{2}{f}+\frac{1}{f_{m}}
\end{aligned}
$$
where, \(f=\) focal length of lens before silvering
\(f_{m}=\) local length of spherical mirror.
$$ \frac{1}{F}=\frac{2}{f}+\frac{2}{R} \quad\left[\because R=2 f_{m}\right] \ldots \text { (i) } $$ Now, \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) Here, \(R_{1}=\infty, R_{2}=30 \mathrm{~cm}\) \(\therefore \quad \frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{30}\right)\) \(\therefore \quad \frac{1}{f}=-\frac{0.5}{30}=-\frac{1}{60}\) or \(\quad f=-60 \mathrm{~cm}\)
...(ii)
Here, \(R_{1}=\infty, R_{2}=30 \mathrm{~cm}\)
\(\therefore \quad \frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{30}\right)\)
or \(\frac{1}{f}=-\frac{0.5}{30}=-\frac{1}{60}\)
or \(f=-60 \mathrm{~cm}\)
Hence, from Eq. (i), we get
$$
\begin{aligned}
&\frac{1}{F}=\frac{2}{60}+\frac{2}{30}=\frac{6}{60} \\
&F=10 \mathrm{~cm}
\end{aligned}
$$
Again given that,
Size of object \(=\) Size of image
$$
\begin{array}{ll}
\text { i.e., } & O=I \\
\therefore & m=-\frac{v}{u}=\frac{1}{0} \\
\Rightarrow \quad & \frac{v}{u}=-1 \\
\text { or } \quad & v=-u
\end{array}
$$
Thus, from lens formula,
$$
\begin{aligned}
\frac{1}{F}=\frac{1}{v}-\frac{1}{u} & \Rightarrow \frac{1}{10}=\frac{1}{-u}-\frac{1}{u} \\
\frac{1}{10} &=-\frac{2}{u} \\
u &=-20 \mathrm{~cm}
\end{aligned}
$$
Hence, to get a real image, object must be placed at a distance 20 cm on the left side of lens.
